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Given that f(x) is a cubic function with zeros at -3, 0, and 7, find an equation for f(x) given that f(-8) = -5.

Sagot :

gmany

Answer:

[tex]\huge\boxed{f(x)=\dfrac{1}{120}x^3-\dfrac{1}{30}x^2-\dfrac{7}{40}x}[/tex]

Step-by-step explanation:

[tex]f(x)=a(x+3)(x-0)(x-7)=ax(x+3)(x-7)\\\\=ax\bigg((x)(x)+(x)(-7)+(3)(x)+(3)(-7)\bigg)\\\\=ax(x^2-7x+3x-21)=ax(x^2-4x-21)\\\\=(ax)(x^2)+(ax)(-4x)+(ax)(-21)\\\\=ax^3-4ax^2-21ax\qquad(*)\\\\f(-8)=-5[/tex]

[tex]\text{substitute}\ x=-8\\\\f(-8)=(a)(-8)^3-(4a)(-8)^2-(21)(-8)a=-512a-(4a)(64)+168a\\\\=-512a-256a+168a=-600a\\\\f(-8)=-5\\\\\text{therefore}\\\\-600a=-5\qquad|\text{divide both sides by (-600)}\\\\a=\dfrac{-5}{-600}\\\\a=\dfrac{1}{120}[/tex]

[tex]\text{Substitute to}\ (*):\\\\f(x)=\dfrac{1}{120}x^3-4\cdot\dfrac{1}{120}x^2-21\cdot\dfrac{1}{120}x=\dfrac{1}{120}x^3-\dfrac{1}{30}x^2-\dfrac{7}{40}x[/tex]

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